\documentclass[UTF8]{ctexart}
\usepackage{amsmath}
\title{矩阵应用与分析作业}
\author{黎吉国&201618013229046}
\begin{document}
\maketitle
\section{第三题}
求下列齐次方程的通解。\\
(a)\\
\begin{eqnarray}
  x_1+2x_2+x_3+2x_4&=&0 \nonumber \\
  2x_1+4x_2+x_3+3x_4&=&0 \nonumber \\
  3x_1+6x_2+x_3+4x_4&=&0 \nonumber
\end{eqnarray}
解：\\
该齐次方程的增广矩阵为:\\
\begin{equation*}
  [A|b]=\left(
  \begin{array}{llll}
    1&2&1&2 \\
    2&4&1&3 \\
    3&6&1&4 \\
  \end{array}
  \middle |
  \begin{array}{l}
    0 \\
    0 \\
    0 \\
  \end{array}
  \right)
\end{equation*}
将其化为行阶梯标准型：\\
\begin{equation*}
  [A|b]=\left(
  \begin{array}{llll}
    1&2&1&2 \\
    2&4&1&3 \\
    3&6&1&4 \\
  \end{array}
  \middle |
  \begin{array}{l}
    0 \\
    0 \\
    0 \\
  \end{array}
  \right)
  \to\left(
  \begin{array}{llll}
    1&2&0&1 \\
    0&0&1&1 \\
    0&0&0&0 \\
  \end{array}
  \middle |
  \begin{array}{l}
    0 \\
    0 \\
    0 \\
  \end{array}
  \right)
\end{equation*}
可得：\\
\begin{align}
  x_1&=-2x_2-x_4 \nonumber \\
  x_2& \text{ is free} \nonumber \\
  x_3&=-x_4 \nonumber \\
  x_4& \text{ is free}\nonumber
\end{align}
它的通解为：\\
\begin{equation*}
X=\left(
\begin{array}{c}
  x_1 \\
  x_2 \\
  x_3 \\
  x_4
\end{array}
\right)
=\left(
\begin{array}{c}
-2x_2-x_4 \\
x_2 \\
-x_4 \\
x_4
\end{array}
\right)
= x_2 \left(
\begin{array}{c}
-2 \\
1 \\
0 \\
0
\end{array}
\right)
+x_4\left(
\begin{array}{c}
-1 \\
0 \\
-1 \\
1
\end{array}
\right)
\end{equation*}

(b)\\
\begin{eqnarray}
  2x+y+z&=&0 \nonumber \\
  4x+2y+z&=&0 \nonumber \\
  6x+3y+z&=&0 \nonumber \\
  8x+4y+z&=&0 \nonumber
\end{eqnarray}
解：\\
该齐次方程的增广矩阵为：\\
\begin{equation*}
  [A|b]=\left(
  \begin{array}{ccc}
    2&1&1 \\
    4&2&1 \\
    6&3&1 \\
    8&4&1 \\
  \end{array}
  \middle |
  \begin{array}{c}
    0 \\
    0 \\
    0 \\
    0
  \end{array}
  \right)
\end{equation*}
将其化为行阶梯标准型为：\\
\begin{equation*}
  [A|b]=\left(
  \begin{array}{ccc}
    2&1&1 \\
    4&2&1 \\
    6&3&1 \\
    8&4&1 \\
  \end{array}
  \middle |
  \begin{array}{c}
    0 \\
    0 \\
    0 \\
    0
  \end{array}
  \right)
  \to \left(
  \begin{array}{ccc}
    2&1&1 \\
    0&0&-1 \\
    0&0&-2 \\
    0&0&-3
  \end{array}
  \middle |
  \begin{array}{c}
    0 \\
    0 \\
    0 \\
    0
  \end{array}
  \right)
  \to \left(
  \begin{array}{ccc}
    1&0.5&0 \\
    0&0&1 \\
    0&0&1 \\
    0&0&1
  \end{array}
  \middle |
  \begin{array}{c}
    0 \\
    0 \\
    0 \\
    0
  \end{array}
  \right)
\end{equation*}
可得：\\
\begin{align}
x&=-0.5y  \nonumber \\
y& \text{ is free}  \nonumber \\
z&=0 \nonumber
\end{align}
该齐次方程的通解为：\\
\begin{equation*}
\left(
  \begin{array}{c}
    x \\
    y \\
    z
  \end{array}
\right)
=y
\left(
  \begin{array}{c}
    -0.5 \\
    1 \\
    0
  \end{array}
\right)
\end{equation*}

\section{第四题}
求下列非齐次方程的通解。\\
(a)\\
\begin{eqnarray}
  x_1+2x_2+x_3+2x_4&=&3 \nonumber \\
  2x_1+4x_2+x_3+3x_4&=&4 \nonumber \\
  3x_1+6x_2+x_3+4x_4&=&5 \nonumber
\end{eqnarray}
解：\\
该非齐次方程的增广矩阵为:\\
\begin{equation*}
  [A|b]=\left(
    \begin{array}{cccc}
      1&2&1&2 \\
      2&4&1&3 \\
      3&6&1&4
    \end{array}
    \middle |
    \begin{array}{c}
      3 \\
      4 \\
      5
    \end{array}
  \right)
\end{equation*}
将该增广矩阵化为行阶梯标准型为：\\
\begin{equation*}
  [A|b]=\left(
    \begin{array}{cccc}
      1&2&1&2 \\
      2&4&1&3 \\
      3&6&1&4
    \end{array}
    \middle |
    \begin{array}{c}
      3 \\
      4 \\
      5
    \end{array}
  \right)
  \to \left(
  \begin{array}{cccc}
    1&2&1&2 \\
    0&0&-1&-1 \\
    0&0&-2&-2 \\
  \end{array}
  \middle |
  \begin{array}{c}
    3 \\
    -2 \\
    4
  \end{array}
  \right)
  \to \left(
  \begin{array}{cccc}
    1&2&0&1 \\
    0&0&1&1 \\
    0&0&0&0
  \end{array}
  \middle |
  \begin{array}{c}
    1 \\
    2 \\
    0
  \end{array}
  \right)
\end{equation*}
可得：\\
\begin{align}
x_1 &= 1-2x_2-x_4 \nonumber \\
x_2 &\text{ is free} \nonumber \\
x_3 &=2-x_4 \nonumber \\
x_4 &\text{ is free}\nonumber
\end{align}
可得这个非齐次方程的通解为：\\
\begin{equation*}
X=\left(
\begin{array}{c}
  x_1 \\
  x_2 \\
  x_3 \\
  x_4
\end{array}
\right)
=\left(
\begin{array}{c}
  1 \\
  0 \\
  2 \\
  0
\end{array}
\right)
+x_2\left(
\begin{array}{c}
  -2 \\
  1 \\
  0 \\
  0
\end{array}
\right)
+x_4\left(
\begin{array}{c}
  -1 \\
  0 \\
  -1 \\
  1
\end{array}
\right)

\end{equation*}
(b)\\
\begin{eqnarray}
  2x+y+z&=&4 \nonumber \\
  4x+2y+z&=&6 \nonumber \\
  6x+3y+z&=&8 \nonumber \\
  8x+4y+z&=&10 \nonumber
\end{eqnarray}
解：\\
该非齐次方程的增广矩阵为：\\
\begin{equation*}
  [A|b]=\left(
    \begin{array}{ccc}
      2&1&1 \\
      4&2&1 \\
      6&3&1 \\
      8&4&1
    \end{array}
    \middle |
    \begin{array}{c}
      4 \\
      6 \\
      8 \\
      10
    \end{array}
  \right)
\end{equation*}
将该增广矩阵化为行阶梯标准型：\\
\begin{equation*}
  [A|b]=\left(
    \begin{array}{ccc}
      2&1&1 \\
      4&2&1 \\
      6&3&1 \\
      8&4&1
    \end{array}
    \middle |
    \begin{array}{c}
      4 \\
      6 \\
      8 \\
      10
    \end{array}
  \right)
  \to\left(
    \begin{array}{ccc}
      1&0.5&0 \\
      0&0&1 \\
      0&0&0 \\
      8&0&0
    \end{array}
    \middle |
    \begin{array}{c}
      2 \\
      2 \\
      0 \\
      0
    \end{array}
  \right)
\end{equation*}
可得：\\
\begin{align}
x &=1-0.5y \nonumber \\
y &\text{ is free} \nonumber \\
z &=2 \nonumber
\end{align}
该非齐次方程的通解为：\\
\begin{equation*}
\left(
\begin{array}{c}
  x \\
  y \\
  z
\end{array}
\right)
=\left(
\begin{array}{c}
  1 \\
  0 \\
  2
\end{array}
\right)
+y\left(
\begin{array}{c}
  -0.5 \\
  1 \\
  0
\end{array}
\right)
\end{equation*}
\end{document}
